$$$\sqrt{2} \sqrt{x} - x^{2}$$$ 的積分

求$$$\int \left(\sqrt{2} \sqrt{x} - x^{2}\right)\, dx$$$。

逐項積分：

$${\color{red}{\int{\left(\sqrt{2} \sqrt{x} - x^{2}\right)d x}}} = {\color{red}{\left(- \int{x^{2} d x} + \int{\sqrt{2} \sqrt{x} d x}\right)}}$$

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：

$$\int{\sqrt{2} \sqrt{x} d x} - {\color{red}{\int{x^{2} d x}}}=\int{\sqrt{2} \sqrt{x} d x} - {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\int{\sqrt{2} \sqrt{x} d x} - {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\sqrt{2}$$$ 與 $$$f{\left(x \right)} = \sqrt{x}$$$：

$$- \frac{x^{3}}{3} + {\color{red}{\int{\sqrt{2} \sqrt{x} d x}}} = - \frac{x^{3}}{3} + {\color{red}{\sqrt{2} \int{\sqrt{x} d x}}}$$

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=\frac{1}{2}$$$：

$$- \frac{x^{3}}{3} + \sqrt{2} {\color{red}{\int{\sqrt{x} d x}}}=- \frac{x^{3}}{3} + \sqrt{2} {\color{red}{\int{x^{\frac{1}{2}} d x}}}=- \frac{x^{3}}{3} + \sqrt{2} {\color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=- \frac{x^{3}}{3} + \sqrt{2} {\color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}$$

因此，

$$\int{\left(\sqrt{2} \sqrt{x} - x^{2}\right)d x} = \frac{2 \sqrt{2} x^{\frac{3}{2}}}{3} - \frac{x^{3}}{3}$$

加上積分常數：

$$\int{\left(\sqrt{2} \sqrt{x} - x^{2}\right)d x} = \frac{2 \sqrt{2} x^{\frac{3}{2}}}{3} - \frac{x^{3}}{3}+C$$

$$$\int \left(\sqrt{2} \sqrt{x} - x^{2}\right)\, dx = \left(\frac{2 \sqrt{2} x^{\frac{3}{2}}}{3} - \frac{x^{3}}{3}\right) + C$$$A